When 4.5 mol Al reacts with 11.5 mol HCl,what is the limiting reactant, and how many moles of H2 can be formed?
LR = limiting reagent.
2Al + 6HCl ==> 3H2 + 2AlCl3
How many mols H2 produced if we had 4.5 mol Al and all of the HCl we needed. That would be
4.5 x (3 mols H2/2 mol Al) = 4.5 x 3/2 = 6.75.
How many mols H2 produced if we had 11.5 mols HCl and all of the Al we needed. That's
11.5 x (3 mols H2/6) = 11.5 x 3/6 = 4.75 mols H2.
Obviously, both answers can't be right; in LR problems the correct answer is ALWAYS the SMALLER value AND the reagent providing that smaller value is the LR. So you will have 5.75 mols H2 produced.
To determine the limiting reactant and the moles of H2 formed, we need to compare the stoichiometry of the reaction to the amounts of reactants given.
The balanced chemical equation for the reaction between aluminum (Al) and hydrochloric acid (HCl) is:
2 Al + 6 HCl -> 2 AlCl3 + 3 H2
Given:
- Moles of Al = 4.5 mol
- Moles of HCl = 11.5 mol
First, we need to establish which reactant will run out first, known as the limiting reactant. We can do this by comparing the moles of each reactant to their stoichiometric ratios in the balanced equation.
For Al:
From the balanced equation, we see that 2 moles of Al react with 6 moles of HCl to produce 3 moles of H2. This gives us a ratio of 2:6:3.
Dividing the moles of Al by its stoichiometric coefficient (2), we get:
Moles of Al available = 4.5 mol / 2 = 2.25 mol
For HCl:
Dividing the moles of HCl by its stoichiometric coefficient (6), we get:
Moles of HCl available = 11.5 mol / 6 ≈ 1.92 mol
Comparing the values obtained, we can see that the number of moles of HCl available is smaller than the moles of Al available. Therefore, HCl is the limiting reactant.
Next, we need to determine the moles of H2 that can be formed using the limiting reactant.
From the balanced equation, we know that for every 6 moles of HCl, 3 moles of H2 are formed.
Since HCl is the limiting reactant, we can use its moles to calculate the moles of H2:
Moles of H2 = (Moles of limiting reactant) * (Moles of H2 / Moles of limiting reactant)
= 1.92 mol * (3 mol H2 / 6 mol HCl)
= 0.96 mol H2
Therefore, when 4.5 mol Al reacts with 11.5 mol HCl, HCl is the limiting reactant, and 0.96 mol of H2 can be formed.